Replacement Model 2

Replacement
Model
2102A71004
2102A71017
2102A71041 BBA 3rd year
2102D70009 IMBA
Replacement Model
All industrial and military equipment gets worn with time and usage and it
functions with decreasing efficiency.
For example : A machine requires higher operating cost, a transport

vehicle such as a car or airplane requires more and more maintenance
cost.
• A railway timetable becomes more and more out of date with the
passage of time.
• The ever increasing repair, maintenance and operating cost
necessitates the replacement of the equipment. However, there is no
sharp, clearly defined time which indicates the need for this
replacement.
The replacement policy, in this case, consists of calculating the increased
operating cost, maintenance cost, forced idle time cost together with cost of
the new equipment and scrap value of the old.
Types of Failures
There are two types of failures:

1. Gradual failure : Gradual failure is progressive in nature. As the life of
the equipment increases, its operational efficiency decreases. This
results in
(a) increased running (repair, maintenance and operating) costs.
(b) decreased productivity.
(c) decreased resale or scrap value.
Machines, vehicles, tyres, tubes, pistons, piston rings, bearings, etc. fall in
this category.
2. Sudden Failure: Some items do not deteriorate with time. They give
the desired level of service for some period, after which they fail. The
period of desired service is not constant but follows some frequency
distribution which may be progressive, retrogressive or random in nature.
(i) Progressive failure: If the probability of failure of an item increases
with increase in its life, then such a failure is called a progressive
failure. Electric bulbs and tubes fall under this category of failure.
(ii) Retrogressive failure: If the probability of failure of an item is more in
the beginning but decreases with the life of an item, then such a failure
is called a retrogressive failure. Automobile engines fall under this
category.
(iii) Random failure: If the probability of failure of the item is due to
random causes such as physical shock, irrespective of its age, then
such a failure is called a random failure. failure of vacuum tubes and
electronic items is generally random in nature
O.R. Methodology of Solving Replacement Problem
OR provides a methodology for tackling replacement problem

which is discussed below:
i) Identify the items to be replaced and also their failure mechanism.
Ii) Collect the data relating to the depreciation cost and the
maintenance cost for the items which follow gradual failure
mechanism. In case of sudden failure of items, collect the data for
replacement cost of the failed items.
Iii) Select a suitable replacement model

Types of Replacement Problems
i) Replacement policy for items, efficiency of which declines gradually

with time without change in money value.
Ii) Replacement policy for items, efficiency of which declines

gradually with time but with change in money value.
Iii) Replacement policy of items breaking down suddenly
a) Individual replacement policy

b) Group replacement policy
Que no. 1 : The cost of a machine is Rs. 6100. and its scrape value is Rs. 100 the maintenance cost per year are as
follows :
Year Maintenance
1 100
2 250
3 400
4 600
5 900
6 1200
7 1600
8 2000
When should be the machine be replaced?

NOTE : If running cost ( Operating and maintenance cost) for the next year {f(n)+1} is
more than the average annual cost nth year then replace at the end of year.
HERE : Cost of machine(C) = 6100, Scrape value(s)=100, n = Number of years,

F(t) ∑ F(t) T=( C – S )+∑ F(t) [ C - S +∑ F(t) ]
Year C–S ( Annual ( Summation of ( Total cost) (Average annual
maintenance maintenance ) cost)
cost )
1 6000 100 100 6100 6100

2 6000 250 100+250=350 6350 3175
3 6000 400 350+400=750 6750 2250
4 6000 600 750+600=1350 7350 1837.5
5 6000 900 1350+900=2250 8250 1650
6 6000 1200 2250+1200=3450 9450 1575
7 6000 1600 3450+1600=5050 11050 1578.5
8 6000 2000 5050+2000=7050 13050 1631.25
we observe from the table that the average annual cost is minimum(Rs.1575) during
the 6th year and then rises . Hence the machine should be replaced after 6th year.
Que no.2 :- A taxi owner estimates from his past record that the cost/year for operating a taxi. where purchase
price when new is Rs. 60000.
Age Operating cost

1 10,000
2 12,000
3 15,000
4 18,000
5 20,000
After fifth year the operating cost is 6000k. Where k=6,7,8,9,10. { k denoting age in years}.
If the resale value decreases by 10% of the purchase price each year what is the best
replacement policy ?
How to calculate the resale value :
10% of purchase price = Rs. 60,000 X10/100 = Rs. 6,000.
Thus the resale value decreases by Rs. 6,000 every year.

which means that ( C – S ) increases by Rs. 6,000 every year.
Age Resale C-S F(t) ∑ F(t) T= C - S+∑ F(t) [ c-s+∑ F(t) ]
value(S)
1 54,000 6,000 10,000 10,000 16,000 16,000
2 48,000 12,000 12,000 22,000 34,000 17,000
3 42,000 18,000 15,000 37,000 55,000 18,333.33
4 36,000 24,000 18,000 55,000 79,000 19,750
5 30,000 30,000 20,000 75,000 1,05,000 21,000
6 24,000 36,000 36,000 1,11,000 1,47,000 24,500
7 18,000 42,000 42,000 1,53,000 1,95,000 27,857.14
8 12,000 48,000 48,000 2,01,000 2,49,000 31,125
9 6,000 54,000 54,000 2,55,000 3,09,000 34,333.33
10 0 60,000 60,000 3,15,000 3,75,000 37,500
Hence 1st year is the optimal replacement period.

Thank
you

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Replacement

For example : A machine requires higher operating cost, a transport

There are two types of failures:

OR provides a methodology for tackling replacement problem

i) Identify the items to be replaced and also their failure mechanism.

Iii) Select a suitable replacement model

i) Replacement policy for items, efficiency of which declines gradually

Ii) Replacement policy for items, efficiency of which declines

Iii) Replacement policy of items breaking down suddenly

a) Individual replacement policy

When should be the machine be replaced?

HERE : Cost of machine(C) = 6100, Scrape value(s)=100, n = Number of years,

1 6000 100 100 6100 6100

Age Operating cost

10% of purchase price = Rs. 60,000 X10/100 = Rs. 6,000.

Thus the resale value decreases by Rs. 6,000 every year.

Hence 1st year is the optimal replacement period.

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